Euler's totient function

Computing Euler's totient function

There are several formulae for computing ${\displaystyle \phi (n)}$.

Euler's product formula

It states $${\displaystyle \phi (n)=n\prod _{p\mid n}\left(1-\frac{1}{p}\right),}$$ where the product is over the distinct prime numbers dividing n.

An equivalent formulation is $${\displaystyle \phi (n)=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)\cdots p_r^{k_r-1}(p_r-1),}$$ where ${\displaystyle n=p_1^{k_1}{p}_2^{k_2}\cdots p_r^{k_r}}$ is the prime factorization of ${\displaystyle n}$ (that is, ${\displaystyle p_1,p_2,\dots ,p_r}$ are distinct prime numbers).

The proof of these formulae depends on two important facts.

Phi is a multiplicative function

This means that if ${\displaystyle gcd(m,n)=1}$, then ${\displaystyle \phi (m)\phi (n)=\phi (mn)}$. Proof outline: Let ${\displaystyle A,B,C}$ be the sets of positive integers which are coprime to and less than m, n, mn, respectively, so that ${\displaystyle |A|=\phi (m)}$, etc. Then there is a bijection between ${\displaystyle A\times B}$ and C by the Chinese remainder theorem.

Value of phi for a prime power argument

If p is prime and ${\displaystyle k\ge 1}$, then

$${\displaystyle \phi \left(p^k\right)=p^k-p^{k-1}=p^{k-1}(p-1)=p^k\left(1-\frac{1}{p}\right).}$$

Proof: Since p is a prime number, the only possible values of ${\displaystyle gcd(p^k,m)}$ are ${\displaystyle 1,p,p^2,\dots ,p^k}$, and the only way to have ${\displaystyle gcd(p^k,m)>1}$ is if m is a multiple of p, that is, ${\displaystyle m\in \{p,2p,3p,\dots ,p^{k-1}p=p^k\}}$, and there are ${\displaystyle p^{k-1}}$ such multiples not greater than ${\displaystyle p^k}$. Therefore, the other ${\displaystyle p^k-p^{k-1}}$ numbers are all relatively prime to ${\displaystyle p^k}$.

Proof of Euler's product formula

The fundamental theorem of arithmetic states that if n > 1 there is a unique expression ${\displaystyle n=p_1^{k_1}{p}_2^{k_2}\cdots p_r^{k_r},}$ where p₁ < p₂ < ... < p_r are prime numbers and each k_i ≥ 1. (The case n = 1 corresponds to the empty product.) Repeatedly using the multiplicative property of φ and the formula for φ(p^k) gives

$${\displaystyle \begin{array}{rcl}\phi (n)& =& \phi (p_1^{k_1})\phi (p_2^{k_2})\cdots \phi (p_r^{k_r})\\ & =& p_1^{k_1}\left(1-\frac{1}{p_1}\right)p_2^{k_2}\left(1-\frac{1}{p_2}\right)\cdots p_r^{k_r}\left(1-\frac{1}{p_r}\right)\\ & =& p_1^{k_1}{p}_2^{k_2}\cdots p_r^{k_r}\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_r}\right)\\ & =& n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_r}\right).\end{array}}$$

This gives both versions of Euler's product formula.

An alternative proof that does not require the multiplicative property instead uses the inclusion-exclusion principle applied to the set ${\displaystyle \{1,2,\dots ,n\}}$, excluding the sets of integers divisible by the prime divisors.

Example

$${\displaystyle \phi (20)=\phi (2^{2}5)=20(1-\frac{1}{2})(1-\frac{1}{5})=20\cdot \frac{1}{2}\cdot \frac{4}{5}=8.}$$

In words: the distinct prime factors of 20 are 2 and 5; half of the twenty integers from 1 to 20 are divisible by 2, leaving ten; a fifth of those are divisible by 5, leaving eight numbers coprime to 20; these are: 1, 3, 7, 9, 11, 13, 17, 19.

The alternative formula uses only integers:$${\displaystyle \phi (20)=\phi (2^2{5}^1)=2^{2-1}(2-1)5^{1-1}(5-1)=2\cdot 1\cdot 1\cdot 4=8.}$$

Fourier transform

The totient is the discrete Fourier transform of the gcd, evaluated at 1. Let

$${\displaystyle \mathcal{F}\{\mathbf{x}\}[m]=\sum _{k=1}^n{x}_k\cdot e^{-2\pi i\frac{mk}{n}}}$$

where x_k = gcd(k,n) for k ∈ {1, ..., n}. Then

$${\displaystyle \phi (n)=\mathcal{F}\{\mathbf{x}\}[1]=\sum _{k=1}^{n}gcd(k,n)e^{-2\pi i\frac{k}{n}}.}$$

The real part of this formula is

$${\displaystyle \phi (n)=\sum _{k=1}^{n}gcd(k,n)\cos \frac{2\pi k}{n}.}$$

For example, using ${\displaystyle \cos \frac{\pi }{5}=\frac{\sqrt{5}+1}{4}}$ and ${\displaystyle \cos \frac{2\pi }{5}=\frac{\sqrt{5}-1}{4}}$:$${\displaystyle \begin{array}{rcl}\phi (10)& =& gcd(1,10)\cos \frac{2\pi }{10}+gcd(2,10)\cos \frac{4\pi }{10}+gcd(3,10)\cos \frac{6\pi }{10}+\cdots +gcd(10,10)\cos \frac{20\pi }{10}\\ & =& 1\cdot (\frac{\sqrt{5}+1}{4})+2\cdot (\frac{\sqrt{5}-1}{4})+1\cdot (-\frac{\sqrt{5}-1}{4})+2\cdot (-\frac{\sqrt{5}+1}{4})+5\cdot (-1)\\ & & +2\cdot (-\frac{\sqrt{5}+1}{4})+1\cdot (-\frac{\sqrt{5}-1}{4})+2\cdot (\frac{\sqrt{5}-1}{4})+1\cdot (\frac{\sqrt{5}+1}{4})+10\cdot (1)\\ & =& 4.\end{array}}$$Unlike the Euler product and the divisor sum formula, this one does not require knowing the factors of n. However, it does involve the calculation of the greatest common divisor of n and every positive integer less than n, which suffices to provide the factorization anyway.

Divisor sum

The property established by Gauss, that

$${\displaystyle \sum _{d\mid n}\phi (d)=n,}$$

where the sum is over all positive divisors d of n, can be proven in several ways. (See Arithmetical function for notational conventions.)

One proof is to note that φ(d) is also equal to the number of possible generators of the cyclic group C_d ; specifically, if C_d = ⟨g⟩ with g^d = 1, then g^k is a generator for every k coprime to d. Since every element of C_n generates a cyclic subgroup, and each subgroup C_d ⊆ C_n is generated by precisely φ(d) elements of C_n, the formula follows. Equivalently, the formula can be derived by the same argument applied to the multiplicative group of the nth roots of unity and the primitive dth roots of unity.

The formula can also be derived from elementary arithmetic. For example, let n = 20 and consider the positive fractions up to 1 with denominator 20: $${\displaystyle \frac{1}{20},\frac{2}{20},\frac{3}{20},\frac{4}{20},\frac{5}{20},\frac{6}{20},\frac{7}{20},\frac{8}{20},\frac{9}{20},\frac{10}{20},\frac{11}{20},\frac{12}{20},\frac{13}{20},\frac{14}{20},\frac{15}{20},\frac{16}{20},\frac{17}{20},\frac{18}{20},\frac{19}{20},\frac{20}{20}.}$$

Put them into lowest terms: $${\displaystyle \frac{1}{20},\frac{1}{10},\frac{3}{20},\frac{1}{5},\frac{1}{4},\frac{3}{10},\frac{7}{20},\frac{2}{5},\frac{9}{20},\frac{1}{2},\frac{11}{20},\frac{3}{5},\frac{13}{20},\frac{7}{10},\frac{3}{4},\frac{4}{5},\frac{17}{20},\frac{9}{10},\frac{19}{20},\frac{1}{1}}$$

These twenty fractions are all the positive ⁠k/d⁠ ≤ 1 whose denominators are the divisors d = 1, 2, 4, 5, 10, 20. The fractions with 20 as denominator are those with numerators relatively prime to 20, namely ⁠1/20⁠, ⁠3/20⁠, ⁠7/20⁠, ⁠9/20⁠, ⁠11/20⁠, ⁠13/20⁠, ⁠17/20⁠, ⁠19/20⁠; by definition this is φ(20) fractions. Similarly, there are φ(10) fractions with denominator 10, and φ(5) fractions with denominator 5, etc. Thus the set of twenty fractions is split into subsets of size φ(d) for each d dividing 20. A similar argument applies for any n.

Möbius inversion applied to the divisor sum formula gives $${\displaystyle \phi (n)=\sum _{d\mid n}\mu \left(d\right)\cdot \frac{n}{d}=n\sum _{d\mid n}\frac{\mu (d)}{d},}$$

where μ is the Möbius function, the multiplicative function defined by ${\displaystyle \mu (p)=-1}$ and ${\displaystyle \mu (p^k)=0}$ for each prime p and k ≥ 2. This formula may also be derived from the product formula by multiplying out $\prod _{p\mid n}(1-\frac{1}{p})$ to get $\sum _{d\mid n}\frac{\mu (d)}{d}.$

An example:$${\displaystyle \begin{array}{cc}\phi (20)& =\mu (1)\cdot 20+\mu (2)\cdot 10+\mu (4)\cdot 5+\mu (5)\cdot 4+\mu (10)\cdot 2+\mu (20)\cdot 1\\ & =1\cdot 20-1\cdot 10+0\cdot 5-1\cdot 4+1\cdot 2+0\cdot 1=8.\end{array}}$$

Some values

The first 100 values (sequence A000010 in the OEIS) are shown in the table and graph below:

In the graph at right the top line y = n − 1 is an upper bound valid for all n other than one, and attained if and only if n is a prime number. A simple lower bound is ${\displaystyle \phi (n)\ge \sqrt{n/2}}$, which is rather loose: in fact, the lower limit of the graph is proportional to ⁠n/log log n⁠.

Euler's theorem

This states that if a and n are relatively prime then

$${\displaystyle a^{\phi (n)}\equiv 1\mathrm{mod}n.}$$

The special case where n is prime is known as Fermat's little theorem.

This follows from Lagrange's theorem and the fact that φ(n) is the order of the multiplicative group of integers modulo n.

The RSA cryptosystem is based on this theorem: it implies that the inverse of the function a ↦ a^e mod n, where e is the (public) encryption exponent, is the function b ↦ b^d mod n, where d, the (private) decryption exponent, is the multiplicative inverse of e modulo φ(n). The difficulty of computing φ(n) without knowing the factorization of n is thus the difficulty of computing d: this is known as the RSA problem which can be solved by factoring n. The owner of the private key knows the factorization, since an RSA private key is constructed by choosing n as the product of two (randomly chosen) large primes p and q. Only n is publicly disclosed, and given the difficulty to factor large numbers we have the guarantee that no one else knows the factorization.

Other formulae

• ${\displaystyle a\mid b⟹\phi (a)\mid \phi (b)}$
• ${\displaystyle m\mid \phi (a^m-1)}$
• ${\displaystyle \phi (mn)=\phi (m)\phi (n)\cdot \frac{d}{\phi (d)}\text{where }d=\gcd (m,n)}$
  • In particular:
• ${\displaystyle \phi (2m)=\{\begin{array}{ll}2\phi (m)& \text{ if }m\text{ is even}\\ \phi (m)& \text{ if }m\text{ is odd}\end{array}}$
• ${\displaystyle \phi \left(n^m\right)=n^{m-1}\phi (n)}$
• ${\displaystyle \phi (\mathrm{lcm}(m,n))\cdot \phi (\gcd (m,n))=\phi (m)\cdot \phi (n)}$

Compare this to the formula $\mathrm{lcm}(m,n)\cdot \gcd (m,n)=m\cdot n$ (see least common multiple).

• φ(n) is even for n ≥ 3.
  Moreover, if n has r distinct odd prime factors, 2^r | φ(n)
• For any a > 1 and n > 6 such that 4 ∤ n there exists an l ≥ 2n such that l | φ(aⁿ − 1).
• ${\displaystyle \frac{\phi (n)}{n}=\frac{\phi (\mathrm{rad}(n))}{\mathrm{rad}(n)}}$

where rad(n) is the radical of n (the product of all distinct primes dividing n).

• ${\displaystyle \sum _{d\mid n}\frac{{\mu }^2(d)}{\phi (d)}=\frac{n}{\phi (n)}}$
• ${\displaystyle \sum _{\genfrac{}{}{0ex}{}{1\le k\le n-1}{gcd(k,n)=1}}k=\frac{1}{2}n\phi (n)\text{for }n>1}$
• ${\displaystyle \sum _{k=1}^n\phi (k)=\frac{1}{2}\left(1+\sum _{k=1}^n\mu (k){\lfloor \frac{n}{k}\rfloor }^2\right)=\frac{3}{{\pi }^2}{n}^2+O\left(n(\log n{)}^{\frac{2}{3}}(\log \log n{)}^{\frac{4}{3}}\right)}$ ( cited in)
• ${\displaystyle \sum _{k=1}^n\phi (k)=\frac{3}{{\pi }^2}{n}^2+O\left(n(\log n{)}^{\frac{2}{3}}(\log \log n{)}^{\frac{1}{3}}\right)}$ [Liu (2016)]
• ${\displaystyle \sum _{k=1}^n\frac{\phi (k)}{k}=\sum _{k=1}^n\frac{\mu (k)}{k}\lfloor \frac{n}{k}\rfloor =\frac{6}{{\pi }^2}n+O\left((\log n{)}^{\frac{2}{3}}(\log \log n{)}^{\frac{4}{3}}\right)}$
• ${\displaystyle \sum _{k=1}^n\frac{\phi (k)}{k^2}=\frac{6}{{\pi }^2}\log n+\frac{6\gamma }{{\pi }^2}-\frac{{\zeta }^{\prime }(2)}{\zeta (2{)}^2}+O\left(\frac{\log n}{n}\right)}$
• ${\displaystyle \sum _{k=1}^n\frac{k}{\phi (k)}=\frac{315\zeta (3)}{2{\pi }^4}n-\frac{\log n}{2}+O\left((\log n{)}^{\frac{2}{3}}\right)}$
• ${\displaystyle \sum _{k=1}^n\frac{1}{\phi (k)}=\frac{315\zeta (3)}{2{\pi }^4}\left(\log n+\gamma -\sum _{p\text{ prime}}\frac{\log p}{p^2-p+1}\right)+O\left(\frac{(\log n{)}^{\frac{2}{3}}}{n}\right)}$
  (where γ is the Euler–Mascheroni constant).

Menon's identity

In 1965 P. Kesava Menon proved $${\displaystyle \sum _{\stackrel{1\le k\le n}{gcd(k,n)=1}}gcd(k-1,n)=\phi (n)d(n),}$$ where d(n) = σ₀(n) is the number of divisors of n.

Divisibility by any fixed positive integer

The following property, which is unpublished as a specific result but has long been known, has important consequences. For instance it rules out uniform distribution of the values of ${\displaystyle \phi (n)}$ in the arithmetic progressions modulo ${\displaystyle q}$ for any integer ${\displaystyle q>1}$.

• For every fixed positive integer ${\displaystyle q}$, the relation ${\displaystyle q|\phi (n)}$ holds for almost all ${\displaystyle n}$, meaning for all but ${\displaystyle o(x)}$ values of ${\displaystyle n\le x}$ as ${\displaystyle x\to \mathrm{\infty }}$.

This is an elementary consequence of the fact that the sum of the reciprocals of the primes congruent to 1 modulo ${\displaystyle q}$ diverges, which itself is a corollary of the proof of Dirichlet's theorem on arithmetic progressions.

Generating functions

The Dirichlet series for φ(n) may be written in terms of the Riemann zeta function as: $${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\phi (n)}{n^s}=\frac{\zeta (s-1)}{\zeta (s)}}$$ where the left-hand side converges for ${\displaystyle \mathrm{\Re }(s)>2}$.

The Lambert series generating function is

$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\phi (n)q^n}{1-q^n}=\frac{q}{(1-q{)}^2}}$$

which converges for |q| < 1.

Both of these are proved by elementary series manipulations and the formulae for φ(n).

Growth rate

In the words of Hardy & Wright, the order of φ(n) is "always 'nearly n'."

First

$${\displaystyle limsup\frac{\phi (n)}{n}=1,}$$

but as n goes to infinity, for all δ > 0

$${\displaystyle \frac{\phi (n)}{n^{1-\delta }}\to \mathrm{\infty }.}$$

These two formulae can be proved by using little more than the formulae for φ(n) and the divisor sum function σ(n).

In fact, during the proof of the second formula, the inequality

$${\displaystyle \frac{6}{{\pi }^2}<\frac{\phi (n)\sigma (n)}{n^2}<1,}$$

true for n > 1, is proved.

We also have

$${\displaystyle liminf\frac{\phi (n)}{n}\log \log n=e^{-\gamma }.}$$

Here γ is Euler's constant, γ = 0.577215665..., so e^γ = 1.7810724... and e^−γ = 0.56145948....

Proving this does not quite require the prime number theorem. Since log log n goes to infinity, this formula shows that

$${\displaystyle liminf\frac{\phi (n)}{n}=0.}$$

In fact, more is true.

$${\displaystyle \phi (n)>\frac{n}{e^{\gamma }\log \log n+\frac{3}{\log \log n}}\text{for }n>2}$$

and

$${\displaystyle \phi (n)<\frac{n}{e^{\gamma }\log \log n}\text{for infinitely many }n.}$$

The second inequality was shown by Jean-Louis Nicolas. Ribenboim says "The method of proof is interesting, in that the inequality is shown first under the assumption that the Riemann hypothesis is true, secondly under the contrary assumption."

For the average order, we have

$${\displaystyle \phi (1)+\phi (2)+\cdots +\phi (n)=\frac{3n^2}{{\pi }^2}+O\left(n(\log n{)}^{\frac{2}{3}}(\log \log n{)}^{\frac{4}{3}}\right)\text{as }n\to \mathrm{\infty },}$$ due to Arnold Walfisz, its proof exploiting estimates on exponential sums due to I. M. Vinogradov and N. M. Korobov. By a combination of van der Corput's and Vinogradov's methods, H.-Q. Liu (On Euler's function.Proc. Roy. Soc. Edinburgh Sect. A 146 (2016), no. 4, 769–775) improved the error term to $${\displaystyle O\left(n(\log n{)}^{\frac{2}{3}}(\log \log n{)}^{\frac{1}{3}}\right)}$$ (this is currently the best known estimate of this type). The "Big O" stands for a quantity that is bounded by a constant times the function of n inside the parentheses (which is small compared to n²).

This result can be used to prove that the probability of two randomly chosen numbers being relatively prime is ⁠6/π²⁠.

Ratio of consecutive values

In 1950 Somayajulu proved

$${\displaystyle \begin{array}{cc}liminf\frac{\phi (n+1)}{\phi (n)}& =0\text{and}\\ limsup\frac{\phi (n+1)}{\phi (n)}& =\mathrm{\infty }.\end{array}}$$

In 1954 Schinzel and Sierpiński strengthened this, proving that the set

$${\displaystyle \left\{\frac{\phi (n+1)}{\phi (n)},n=1,2,\dots \right\}}$$

is dense in the positive real numbers. They also proved that the set

$${\displaystyle \left\{\frac{\phi (n)}{n},n=1,2,\dots \right\}}$$

is dense in the interval (0,1).

Totient number

A totient number is a value of Euler's totient function: that is, an m for which there is at least one n for which φ(n) = m. The valency or multiplicity of a totient number m is the number of solutions to this equation. A nontotient is a natural number which is not a totient number. Every odd integer exceeding 1 is trivially a nontotient. There are also infinitely many even nontotients, and indeed every positive integer has a multiple which is an even nontotient.

The first few totient numbers are ${\displaystyle 1,2,4,6,8,10,12,16,18,20}$, see sequence A002202.

The number of totient numbers up to a given limit x is

$${\displaystyle \frac{x}{\log x}{e}^{(C+o(1))(\log \log \log x{)}^2}}$$

for a constant C = 0.8178146....

If counted accordingly to multiplicity, the number of totient numbers up to a given limit x is

$${\displaystyle |\{n:\phi (n)\le x\}|=\frac{\zeta (2)\zeta (3)}{\zeta (6)}\cdot x+R(x)}$$

where the error term R is of order at most ⁠x/(log x)^k⁠ for any positive k.

It is known that the multiplicity of m exceeds m^δ infinitely often for any δ < 0.55655.

Ford's theorem

Ford (1999) proved that for every integer k ≥ 2 there is a totient number m of multiplicity k: that is, for which the equation φ(n) = m has exactly k solutions; this result had previously been conjectured by Wacław Sierpiński, and it had been obtained as a consequence of Schinzel's hypothesis H. Indeed, each multiplicity that occurs, does so infinitely often.

However, no number m is known with multiplicity k = 1. Carmichael's totient function conjecture is the statement that there is no such m.

Perfect totient numbers

A perfect totient number is an integer that is equal to the sum of its iterated totients. That is, we apply the totient function to a number n, apply it again to the resulting totient, and so on, until the number 1 is reached, and add together the resulting sequence of numbers; if the sum equals n, then n is a perfect totient number.

Applications

Cyclotomy

In the last section of the Disquisitiones Gauss proves that a regular n-gon can be constructed with straightedge and compass if φ(n) is a power of 2. If n is a power of an odd prime number the formula for the totient says its totient can be a power of two only if n is a first power and n − 1 is a power of 2. The primes that are one more than a power of 2 are called Fermat primes, and only five are known: 3, 5, 17, 257, and 65537. Fermat and Gauss knew of these. Nobody has been able to prove whether there are any more.

Thus, a regular n-gon has a straightedge-and-compass construction if n is a product of distinct Fermat primes and any power of 2. The first few such n are

2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40,... (sequence A003401 in the OEIS).

Prime number theorem for arithmetic progressions

The RSA cryptosystem

Setting up an RSA cryptosystem involves choosing large prime numbers p and q, computing n = pq and k = φ(n), and finding two numbers e and d such that ed ≡ 1 (mod k). The numbers n and e (the "encryption key") are released to the public, and d (the "decryption key") is kept private.

A message, represented by an integer m, where 0 < m < n, is encrypted by computing S = m^e (mod n).

It is decrypted by computing t = S^d (mod n). Euler's Theorem can be used to show that if 0 < t < n, then t = m.

The security of an RSA system would be compromised if the number n could be efficiently factored or if φ(n) could be efficiently computed without factoring n.

Unsolved problems

Lehmer's conjecture

If p is prime, then φ(p) = p − 1. In 1932 D. H. Lehmer asked if there are any composite numbers n such that φ(n) divides n − 1. None are known.

In 1933 he proved that if any such n exists, it must be odd, square-free, and divisible by at least seven primes (i.e. ω(n) ≥ 7). In 1980 Cohen and Hagis proved that n > 10²⁰ and that ω(n) ≥ 14. Further, Hagis showed that if 3 divides n then n > 10^1937042 and ω(n) ≥ 298848.

Carmichael's conjecture

This states that there is no number ${\displaystyle n}$ with the property that for all other numbers ${\displaystyle m}$, ${\displaystyle m\ne n}$, ${\displaystyle \phi (m)\ne \phi (n)}$. See Ford's theorem above.

If there is a single counterexample to this conjecture, there must be infinitely many counterexamples, and the smallest one has at least ten billion digits in base 10.

Riemann hypothesis

The Riemann hypothesis is true if and only if the inequality $${\displaystyle \frac{n}{\phi (n)}<e^{\gamma }\log \log n+\frac{e^{\gamma }(4+\gamma -\log 4\pi )}{\sqrt{\log n}}}$$ is true for all ${\displaystyle n\ge p_{120569}\mathrm{\#}}$ where ${\displaystyle \gamma }$ is Euler's constant and ${\displaystyle p_{120569}\mathrm{\#}}$ is the product of the first 120569 primes.